Answer: B

Small Hint:

Use area ratios from bases on the same line

Big Hint:

Let $x = [EBF]$ and use $[AFD] : [DFC] = 1 : 2$

Video solution:

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Written solution:

Since $AD : DC = 1 : 2,$ triangles $ABD$ and $DBC$ have areas in the ratio $1 : 2.$ Thus $[ABD] = 120$ and $[DBC] = 240.$

Because $E$ is the midpoint of $BD,$ triangles $ABE$ and $AED$ each have area $60.$ Let $x = [EBF].$ Then $[DEF] = x$ as well, since $BE = ED$ and both triangles have their third vertex on line $AF.$

Segment $DF$ splits $\triangle DBC,$ so $[DFC] = 240 - 2x.$ Also, $\triangle ADF$ and $\triangle DFC$ have bases $AD$ and $DC$ on the same line, so their areas are in the ratio $1 : 2.$

Therefore $$\dfrac{60 + x}{240 - 2x} = \dfrac{1}{2}.$$ Solving gives $120 + 2x = 240 - 2x,$ so $x = 30.$

Thus, the correct answer is B.